Question

x⁵-4x³ + x²-4=0

(3x-5)(2x-5)-x²>2x-3

Prosze o dokladny opis krok po kroku co robimy.Dzieki!

Answer 1

1)

x⁵-4x³+x²-4=0

x³(x²-4)+1(x²-4)=0

(x³+1)(x²-4)=0

(x+1)(x²-x+1)(x²-4)=0

(x+1)(x²-x+1)(x+2)(x-2)=0

x+1=0

x₁=-1          <---

 

x²-x+1=0

Δ=1-4=-3

sprzeczność

 

x+2=0

x₂=-2          <---

 

x-2=0

x₃=2          <---

 

2)

(3x-5)(2x-5)-x²>2x-3

6x²-15x-10x+25-x²>2x-3

5x²-25x+25-2x+3>0

5x²-27x+28>0

Δ=(-27)²-4·5·28=729-560=169

√Δ=13

x₁=(27-13):10=14:10=1,4

x₂=(27+13):10=40:10=4

x∈(-∞;1,4)u(4;+∞)          <---